Particle on a fixed plane curve ($(1+y^{\prime 2})\ddot{x}=-g{y}^{\prime }-y^{\prime }{y}^{\prime \prime }{\dot{x}}^2$)

A point mass constrained to the fixed graph $y=y(x)$ has velocity $(\dot{x},y^{\prime }\dot{x})$. Under uniform downward gravity, its Lagrangian is $L=m(1+y^{\prime 2}){\dot{x}}^2/2-mgy$, so its canonical momentum is $p_x=m(1+y^{\prime 2})\dot{x}$, not the horizontal mechanical momentum. The Euler-Lagrange equation gives the equation in the title, and the conserved energy is $m(1+y^{\prime 2}){\dot{x}}^2/2+mgy$. A straight segment with constant slope $s$ has constant horizontal acceleration $-gs/(1+s^2)$; zero curvature does not imply zero acceleration. A cart held to a track satisfies this mechanical constraint only while the assumed contact or retaining mechanism remains effective.